Assignment 6 Solution.pdf - ASSIGNMENT 6 Due by Friday by...

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ASSIGNMENT 6 Due by Friday 05/31/2019 by 11:59pm 1. Solve the problem given below (6 pts) Solution: 5.3.1: What is the cache block size? Cache block size = 2 offset-bits = 2 5 bytes = 2 3 words = 8 words Cache block size is 8 words long. 5.3.2: How many entries does the cache have? Cache Size = 2 index bits = 2 5 = 32 entries 5.3.3: What is the ratio between total bits required for such a cache implementation over data storage bits Total bits required = 2 index bits * block size * tagsize * valid bit size) = 32* (256+ 22+1) = 32* 279 bits Data Storage bits = 2 index bits * block size = 32 * 32 * 8 = 32 * 256 bits Ratio = 32 * 279/ (32 * 256) = 279 / 256 = 1.0898
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2. Solve problem: (6 pts) 5.9.1 What is the optimal block size for a miss latency of 20 × B cycles? Answer: AMAT = Average Memory Access Time = time for hit + miss rate * latency.
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Unformatted text preview: Block Size Miss Rate Latency AMAT 8 4% 8*20 = 160 1+0.04*160 = 7.4 16 3% 16*20 = 320 1+0.03*320 = 10.6 32 2% 32*20 = 640 1+0.02*640 = 13.8 64 1.5% 64*20 = 1280 1+0.15*1280 = 20.2 128 1% 128*20 = 5120 1+0.01*5120 = 26.6 Optimal block size is 8 as it has low AMAT. 5.9.2 What is the optimal block size for a miss latency of 24 + B cycles? Block Size Miss Rate Latency AMAT 8 4% 8+24 = 32 1+0.04*32 = 2.28 16 3% 16+24 = 40 1+0.03*40 = 2.20 32 2% 32+24 = 56 1+0.02*56= 2.12 64 1.5% 64+24 = 88 1+0.15*88 = 2.32 128 1% 128+24 = 152 1+0.01*152 = 2.52 Optimal block size is 32 as it has low AMAT 5.9.3 For constant miss latency, what is the optimal block size? Answer: Memory stall = miss rate * miss penalty For constant miss latency the optimal block size is 128....
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